Do you know any board or technical information for the following application: Input 110VAC and Output 320 VDC Power 1hp? Let me know your comments. Regards,
Hi Suresh, Electronic Devices is about understanding physics behind the device operation to get IV or CV characteristics. Analog circuits is about using devices characteristics to design circuits for application. If you can draw proper boundaries, understanding devices and characteristics is enough to start Analog. If curiousity levels are high (to avoid certain doubts/questions) completion of devices is recommended.
From output voltage if you remove DC component the waveform is symmetric wrt to time axis. Then peak to peak voltage is Vr. Therefore Voltage peak is Vr/2. I believe its correct
Thank you sir. you have simplified everything and i really appreciate it. Komape from south africa.
Well
Thank u :) I like such deep explanations.
what if Rc
Do you know any board or technical information for the following application: Input 110VAC and Output 320 VDC Power 1hp? Let me know your comments. Regards,
Sir your videos are good but they are underrated, why I don't understand
Very well explained !!
why is it when RLC>> T, the T discharged=T, supposed to be Tcharged=T
Analog start karne se phle edc ki requirement h ya nhi
Hi Suresh,
Electronic Devices is about understanding physics behind the device operation to get IV or CV characteristics.
Analog circuits is about using devices characteristics to design circuits for application.
If you can draw proper boundaries, understanding devices and characteristics is enough to start Analog.
If curiousity levels are high (to avoid certain doubts/questions) completion of devices is recommended.
please give the refference of this topic sir?
Thnks broo
Can someone explain how we subbed Idc there, the relation?
using V=IR relation
Sir, why Vr/2 ??
From output voltage if you remove DC component the waveform is symmetric wrt to time axis. Then peak to peak voltage is Vr. Therefore Voltage peak is Vr/2. I believe its correct
How did e^(-T/RLC) become 1 - (T/RLC)? Since RLC is very big compared to T, wouldn't that mean that it's approaching 0, thus e^0 = 1?
expansion of e^-x = 1- x, when x >T T/RC